An inductor with an inductance of 2.10 H and a resistance of 7.90 Omega is conne

Question

An inductor with an inductance of 2.10 H and a resistance of 7.90 Omega is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance.

Find the initial rate of increase of current in the circuit.

di/dt = _____ A/s

Find the rate of increase of current at the instant when the current is 0.550A

di/dt = _____ A/s

Find the current 0.210 after the circuit is closed.

I = _______ A

Find the final steady-state current.

I = _______ A

Explanation / Answer

V = L di/dt 6.00 V = 2.10 * di/dt di/dt = 2.86 A/s V across resistor when current is 0.550 A: 0.550 * 7.9 = 4.345 V V across inductor = 6 - 4.345 = 1.655 V 1.655 = L di/dt di/dt = 0.788 A/s You can treat this as a series RL problem. the differential equation is L di/dt + iR = E di/dt = 1/L (E - iR) di/ (E - iR) = 1/L dt integrating both sides, you get, (E - i(t)*R)/ (E - i(0) * R) = Rt/L taking natural logs of both sides, and rearranging E - i(t) * R = (E - i(0)*R) e^(-Rt/L) i(0) = 0 rearranging, i(t) = E/R(1-e^-(Rt/L)) i(0.210 s) = 6/7.9 * (1- e^(-0.210 * 7.9/2.1)) = 0.415 A steady state current, inductor is a short, so its just V/R = 6/7.9 = 0.759 A (as t-> infinity, the above equation just becomes E/R)

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